Maths

Unit: 17 Statistics

17.0 Review:

Discuss on groups and find the answer of the following questions.

i. Collect the marks obtained by each members of group in mathematics in
mid-term examination.

ii. Find mean score (mark) and median mark of each group.
iii. Present the result to the class.

iv. Present the scores of all marks and list on the board. Can you calculate
mean and median of the total data as before? Discuss in groups.

To calculate control values from large number of data we have to make frequency
distribution from given data set. We are going to discuss frequency distribution.

17.1 Frequency distribution:

If there are small number of data repeated many times, then we can make the
table of data with respective frequencies of the data series which is called discrete
series.
If the data number is neither small nor repeated, what can we do? Discuss?

In this case we make the suitable class interval and make tally bar for the data
that lie in that interval. At last column, we write the total number of tally in each
interval called frequency.

Let the marks obtained by 40 students in a midterm exam be as follows:

25, 10, 31, 22, 37, 42, 45, 37, 32, 34, 45, 40, 29, 27, 28, 17, 19, 22, 25, 15, 14, 13,
28, 36, 38, 41, 42, 39, 25, 24, 31, 21, 22, 25, 26, 35, 36, 39, 49, 98.

Here, the minimum score is 10, so we make the following frequency distribution
on table as the class interval 10-20 and so on.

Interval Tally bar No of student

10-20 | 6

20-30 llll 14

30-40 lll 13

40-50 ll 7

The process of representation of data by using table is called frequency
distribution. This table is called frequency table and the number of students in

Mathematics, grade 10 221

each class interval is called frequency of that class interval. In each range, the first
data value (number) is called lower limit and the largest data value (number) is
called upper limit of that interval (range).

Example 1:
Construct frequency table of class interval 10 of the following data:
8, 46, 32, 38, 15, 46, 22, 26, 13, 14, 12, 54, 9, 25, 27, 45, 53, 18, 32, 6, 34, 31, 38
Solution:

The minimum value of data is 6, so the first class interval is 0-10 and so on. Then
frequency distribution is

Class Tally Bar frequency
interval

0-10 ||| 3

10-20 5

20-30 |||| 4

30-40 |6

40-50 ||| 3

50-60 || 2

Here, 0 is lower limit and 10 is upper limit of 0-10.
The range of interval = length of interval = 10 – 0 = 10.

Exercise 17.1

1. Construct a frequency table of each of class interval 10 of the following data.
Age of family members (in years)

9, 20, 35, 42, 36, 2, 7, 15, 21, 25, 43, 53, 40, 38, 36, 22, 69, 65, 51, 47, 4, 14, 28,
60, 72, 77, 34, 21, 16, 75, 8, 15, 16, 29, 44
2. Construct a frequency table of class interval 4 of the following data.

weight of 30 students (in kg.)
31, 32, 31, 36, 45, 47, 50, 53, 60, 32, 35, 37, 45, 41, 55, 44, 48, 65, 63, 68, 40, 45,
49, 52, 35, 33, 39, 54, 32
3. The marks obtained by 50 students in a test is given below. Construct the
frequency distribution table of class interval 30-40 with first class.

222 Mathematics, grade 10

Marks obtained by 50 students

74, 62, 71, 63, 79, 73, 35, 43, 49, 48, 56, 59, 32, 35, 72, 58, 57, 62, 38, 49, 45, 42,
44, 43, 48, 52, 56, 72, 64, 39, 48, 62, 77, 44, 39, 75, 79, 83, 84, 81, 66, 69, 35, 44,
30, 83, 77, 44, 55, 48

4. The hourly wages of 36 workers of a factory are given below.
74, 71, 79, 68, 74, 73, 63, 62, 84, 61, 75, 72, 79, 76, 67, 72, 61, 60, 69, 77, 81, 68,
67, 83, 72, 74, 78, 84, 80, 71, 66, 81, 64, 64, 73, 68
Construct the frequency distribution of the above data with class interval of 5.

5. Divide the students in the suitable groups and tell them to collect the age of about
100 neighbors of family members and represent them in frequency distribution
table.

7.2. Central tendency

The value of the given set of date that represent the characteristics of entire data
is called central value. The calculation of such value is called measure of central
tendency. The most common measure of central tendencies is mean, median,
mode and quartiles.

7.2.1 Mean: The mean is the sum of numerical values of each and every observation
divided by the total number of observation. It is denoted by ̅ (x bar) for variable

x.

If , ,....... be n discrete values of variable x then their arithmetic mean is

given by ̅ = ⋯ =∑

If , ,....... be n discrete values with respective frequencies , ,......., ,

their arithmetic mean is given by

̅= ⋯ = ∑ =∑ Where N =∑ = Total frequency.
....... ∑

We have already discussed about those two formulae in our previous classes.

Now we are going to discuss about the mean of grouped or continuous data.

Arithmetic mean for grouped data or continuous series
In grouped data (continuous series), the observations are classified with some suitable
range values along with their class frequencies. To calculate arithmetic mean of grouped
data, first we have to find the mid-value of each interval (range) as shown below

Mid-value (m) =

Mathematics, grade 10 223

Then we can use the formula of calculation of mean of discrete series with placing value
of mid-value (m) instead of value of variable x.

Calculation of mean of grouped data

(a) If , ,....... be mid values of n intervals with frequencies , ,.......,

respectively then the mean is calculated by

̅= ⋯ = ∑ =∑
∑
.......

This method is called direct method of calculation of mean.

Example 1:
Calculate the mean (arithmetic mean) of the data given below.

weight of students (in Kg.) 0-10 10-20 20-30 30-40 40-50 50-60

No. of students 12 18 27 20 17 6

Solution:
Calculation of mean,
Here,

weight (In kg.) No of students (f) mid value (m) f.m
0-10 12 60
10-20 18 0 + 10 270
20-30 27 =5 675
30-40 20 700
40-50 17 2 765
50-60 6 10 + 20 330

= 15
2
20 + 30

= 25
2
30 + 40
2 = 35
40 + 50
2 = 45
50 + 60

= 55
2

= N = 100 = 2800

Now, = 28 kg.
We know, the mean ( ) = ∑ =

Therefore, the mean is 28kg.

224 Mathematics, grade 10

(b) Deviation method (shortcut method)
Let us consider A = assumed mean of the data, then d = m-A
The mean is calculated by ̅ = + ∑

Look at the above example, suppose A = 25

Weight No. of students (f) mid value (m) d = m-A f.d
(in Kg.)

0-10 12 5 -20 -240

10-20 18 15 -10 -180

20-30 27 25 0 0

30-40 20 35 10 200

40-50 17 45 20 340

50-60 6 55 30 180

N = 100 = 300

We know that, mean ( ̅) = + ∑

=25 + = 25 + 3 = 28kg.

(c) Step deviation method: If h be the size of class and A be assumed the mean of

given data set, we can calculate arithmetic mean ̅ as below

̅= +∑ ℩ ℩= and h = mid value of every interval

xh, where

In the above example,
Suppose A = 25 and h = 10, then

weight mid value (m) No. of students ℩ = − 25 f.d'
10
(in Kg.) (f)

0-10 5 12 -2 -24

10-20 15 18 -1 -18

20-30 25 27 0 0

30-40 35 20 1 20

40-50 45 17 2 34

50-60 55 6 3 18

N = 100 ℩ = 30

Mathematics, grade 10 225

We have = +∑ ℩

xh

= 25 + x 10 = 25 + 3 = 28

Example 2:
If ∑ = 2700 and N = 50, find ̅

Solution: = 2700, N = 50, = ?
Here, ∑ = ∑ = = 54

we know,

Example 3:
If assumed mean A = 40, ∑ = 20 and mean ( ̅) = 42, find the value of N.

Solution: =42, N =?

Here, A=40, ∑ = 20,
We know, ̅ = A+∑

Or, 42 = 40 +

Or, = 42-40

Or, 2N = 20
 N = 10

Example 4:
If the mean height of the following data is 157.75 cm, find the value of K.

Height 140-145 145-150 150-155 155-160 160-165 165-170 170-175
(cm)

No of
25 8 k 75 3

Students

Solution:
Calculation of arithmetic mean

Height (in cm) No of students (f) mid value (m) f.m

140-145 2 142.5 285

145-150 5 147.5 737.5

150-155 8 152.5 1220

155-160 k 157.5 157.5k

226 Mathematics, grade 10

Height (in cm) No of students (f) mid value (m) f.m
162.5 1137.5
160-165 7 167.5 837.5
172.5 517.5
165-170 5 ∑ =4735+157.5k

170-175 3

N = 30+k

We know that, = ∑

Or, 157.75 = .

Or, 4732.5+157.75k = 4735+157.5k
Or, 157.75k-157.50k = 4735-4732.5
Or, 0.25k = 2.5
 k = 10

Example 5:
Calculate the mean of the following data by constructing frequency table of class
interval of length 10.
7, 22, 32, 47, 59, 16, 36, 17, 23, 39, 49, 31, 21, 24, 41, 12, 49, 21, 9, 8, 51, 36, 29, 18
Solution:
Construction of frequency table.

Class Tally bar Frequency (f) Mid-value (m) fm
0-10 lll 3 5 15
10-20 llll 4 15 60

20-30 |6 25 150

30-40 llll 5 35 175
40-50 ll 4 45 180
50-60 2 55 110
N = 24 ∑  =690

We have mean X = ∑ 

= =28.75

∴ mean X = 28.75

Mathematics, grade 10 227

Exercise 17.2

1. Find the mean of the following data.
(a) 35, 36, 42, 45, 48, 52, 58, 59
(b) 13.5, 14.2, 15.8, 15.2, 16.9, 16.5, 17.4, 19.3, 15.2
(c)

x 5 8 10 12 14 16

f 4 5 8 10 2 2

(d)

Age ( in yrs) 12 13 14 15 16 17

No. of 2 4 6 12 10 6

students

2. Calculate the mean of the following date by using direct method.
(a)

Age (yrs) 0-10 10-20 20-30 30-40 40-50

No. of children 5 9 15 7 4

(b)

Marks obtained 10-20 20-30 30-40 40-50 50-60 60-70

No. of students 1 4 10 8 7 5

(c)

Daily wages (Rs) 200-400 400-600 600-800 800-1000 1000-1200

No. of workers 3 7 10 6 4

(d)
Class interval 0-10 10-20 20-30 30-40 40-50 50-60

Frequency 75 6 12 8 2

3. Calculate the mean of the Q.(2) by

i. Deviation method/short cut method

ii. Step deviation method

4. Calculate the missing part of the following.

(a) ̅ = 49, ∑ = 980, N =? (b) ̅ = 102.25, N = 8, ∑ =?

228 Mathematics, grade 10

(c) A = 100, ̅ = 90, ∑ =?, N = 10 (d) ̅ = 41.75, ∑ =270, N = 40, A =?

5.(a) If the mean of the given data is 32.5, find the value of k.

Marks obtained 0-10 10-20 20-30 30-40 40-50 50-60

Number of students 5 10 K 35 15 10

(b) If the mean of the following data is 14.2, find the value of P.

X 0-20 20-40 40-60 60-80 80-100

f 35 400 350 p 65

(c) If the mean age of the workers in a factory of the following data is 36.24, find the
value of y.

Age (years) 16-24 24-32 32- 40 40-48 48-56 56-64

No. of workers 6 8 Y 8 4 2

(d) If the average expenditure per week of the students is Rs. 264.67, find the missing
frequency.

Expenditure (Rs.) 0-100 100- 200 200-300 300-400 400-500 500-600

No. of students 20 30 ? 20 18 12

6. Calculate mean of the following data by constructing frequency distribution table.

(a) 15, 51, 32, 12, 32, 33, 23, 43, 35, 46, 57, 19, 59, 25, 20, 38, 16, 45, 39, 40
(construct table of length 10)

(b) 25, 15, 24, 42, 22, 35, 34, 41, 33, 38, 54, 50, 36, 40, 27, 18, 35, 16, 51, 31,
23, 9, 16, 23, 31, 51, 7, 30, 17, 40, 60, 32, 50, 10, 23, 12, 21, 28, 37, 20, 58,
39, 10, 41, 13 (class of length 5)

7. (a) Find mean of the following data

X 0-9 10-19 20-29 30-39 40-49 50-59

F 8 10 14 10 8 10

(b)

Expenditure 0-400 500-900 1000-1400 1500-1900 2000-2400 2400-2800

No. of 1 2 3 4 1 2

workers

8. Divide all students into the groups of 4. Collect the data about their age of at least
50 students of different classes from class 1 to 12, of your school. Construct
the frequency distribution table. Calculate mean by using direct and deviation
method. Prepare a report and present to the class.

Mathematics, grade 10 229

17.2.2 Median:
The central value of a distribution that divides the entire data set into exactly two
equal parts is called median. Median is also called mid-value of the distribution.
The half of the data of the distribution lie below the median value and rest half of
the data are above the median value. It is denoted by Md.
For individual series to calculate the median;
i. Arrange all data in ascending or descending order.

ii. Use the formula ( ) item, where n = total number of data in set.

Similarly, for discrete series the following three steps are used.
iii. Construct cumulative frequency distribution table.
iv. Find , where N = ∑ , sum of frequencies = total numbers of data in data

sets.

v. See cumulative frequency equal or just greater than

vi. Locate the corresponding value of x in the table. which is median value.
Calculation of the median from grouped or continuous data
The following steps are used to calculate median of continuous data or grouped data:
i. Prepare less than cumulative frequency distribution table.

ii. Calculate to find the position of the median.

iii. See the cumulative frequency equal or just greater than and identify the
median class (interval).

iv. Use following formula to find the median value.

Md = L + xh

Where, L = Lower limit of the median class

N = Total Frequency

c.f = Cumulative frequency of class preceding to median class
f = frequency of the median class

h = size of the median class

Example 1:
Calculate the median from the following data.

Mark obtained 15 18 22 26 27 29
10 75
No. of students 348

230 Mathematics, grade 10

Solution: No of students (f) cumulative frequency (cf)
Calculation of median 3 3
4 7
Mark obtained (x) 8 15
15 10 25
18 7 32
22 5 37
26
27 f = N =37
29

We have,

The position of Median (Md) =

= ( ) = 19 item

 The cumulative frequency equal to or greater than 19 is 25. Therefore, the values of
X corresponding to 25 is 26
Therefore, Median (Md) = 26
Example 2:
Calculate the median of the following distribution.

Score of students 0-8 8-16 16-24 24-32 32-40 40-48 48-56

No. of students 6 10 16 18 12 10 8

Solution:
Calculation of median

Score of students (x) No of students (f) Cumulative frequency (c.f)

0-8 6 6

8-16 10 16

16-24 16 32

24-32 18 50

32-40 12 62

40-48 10 72

48-56 8 80

f = N= 80

Mathematics, grade 10 231

Now, the position of median class = ( ) item

= ( ) item

= 40 item
The value of cumulative frequency equal or greater than 40 is 50.
So, median class is 24-32, where L = 24, c.f. = 32, f = 18, h = 8

By formula, Median (Md) = L + xh

= 24 + x8

= 24 +

= 24 + 3.56 = 27.56
Example 3:
Find the missing frequency of the following distribution if the median value is 93.6

X 0-30 30-60 60-90 90-120 120-150 150-180

f 5 p 22 25 14 4

Solution:
Table for calculation of frequency

X f c.f.

0-30 5 5

30-60 p 5+p

60-90 22 27+p

90-120 25 52+p

120-150 14 66+p

150-180 4 70+p

f = N = 70 + p

Given that, Md = 93.6
Md lies in 90-120, where L = 90, f =25, c.f. = 27+p, h =30, N =70+p
Now,

Md = L+ xh

( )( )

Or, 93.6 = 90+ x30

232 Mathematics, grade 10

Or, 93.6-90 = x30

Or, 3.6 = (16 - P)x

Or, 3.6x5 = 48-3p
Or, 3P = 48 -18
Or, 3p = 30
 p = 10
∴ Missing frequency (P) = 10
Example 4:
Find the median height of the plants of the following data.

Height (in cm) 4-6 7-9 10-12 13-15 16-18 19- 21 22-24

No. of plants 2 3 10 7 4 3 2

Solution:

Here the classes are discontinuous. So we need to make them continuous by using
adjustment/continuity/correction factor.

Correction factor =

= = 0.5

The class intervals are made continuous by adding 0.5 in upper limit and subtracting 0.5
in lower limit of each class interval.

Than the table for calculation of median is as follows

Height (cm) No. of plants (f) Cumulative frequency

3.5-6.5 2 2

6.5-9.5 3 5

9.5-12.5 10 15

12.5-15.5 7 22

15.5-18.5 4 26

18.5-21.5 3 29

21.5-24.5 2 31

f = N = 31

Now, the median class = item

= item = 15.5th item

Mathematics, grade 10 233

The value of c.f equal or greater than 15.5 is 22 in the column of cumulative frequency.
So median class is 12.5-15.5 where, L = 12.5, c.f = 15, f = 7 and h = 3

Median (Md) = L+ xh

= 12.5 + . x 3

= 12.5 + . = 12.71

∴ Median height of plant is 12.71 cm

Exercise 17.3

1. Calculate the median from the following data.
(a) 2.5, 4.5, 3.6, 4.9, 5.4, 2.9, 3.1, 4.2, 4.6, 2.2, 1.5
(b) 100, 105, 104, 197, 97, 108, 120, 148, 144, 190, 148, 22, 169, 171, 92, 100
(c)

Marks 18 25 28 29 34 40 44 46

No. of students 3 6 5 7 8 12 5 4

(d)

X 102 105 125 140 170 190 200

f 10 18 22 25 15 12 8

2. Calculate the median from the following frequency distribution table.
(a)

Wt (kg) 30-40 40-50 50-60 60-70 70-80 80-90 90-100

No. of 3 5 7 11 10 3 1
students

(b)

Height 140-145 145-150 150-155 155-160 160-165 165-170 170-175
(cm)

Frequency 2 5 8 10 7 5 3

234 Mathematics, grade 10

(c)

Expenditure 0-100 100-200 200-300 300-400 400-500 500-600
per day (Rs.)

Frequency 22 34 52 20 19 13

(d)

Marks obtained less than 20 40 60 80 100

No. of students 21 44 66 79 90

3. Calculate the missing frequencies in the following table where;
(a) Median (Md) = 35

Mark obtained 20-25 25-30 30-35 35-40 40-45 45-50

No. of students 2 5 8 k 4 5

(b) Median (Md) = 132.5

Wages 100-110 110- 120 120-130 130-140 140-150 150-160

No. of workers 5 6 p 4 75

(c) Median (Md) = 36

Age (yr) 20-25 25-30 30-35 35-40 40-45 45-50 50-55 55-60

No. of persons 50 70 100 300 k 220 70 60

4. Calculate the median of the following data:
(a)

Mark obtained 50-60 60-70 70-80 80-90 90-100
11 16 20
No. of students 2 5
60 80 100
(b) 26 21 16

Mark obtained less than 20 40 800 900 1000
152 177 200
No. of students 14 23

(c)

Income (Rs) less than 600 700

No. of workers 30 98

Mathematics, grade 10 235

(d)

Temp (°c) 0-9 10-19 20-29 30-39 40-49

No. of days 8 10 20 15 7

5.(a) The mark obtained by 40 students of a class in a certain exam is as follow.
Construct a frequency distribution table of class interval of 10 and calculate the
median.

22, 56, 62, 37, 48, 30, 58, 42, 29, 39, 37, 50, 38, 41, 32, 20, 28, 16, 43, 18, 40, 52,
44, 27, 35, 45, 36, 49, 55, 40

(b) The height (in cm) of 40 students of grade X is given below. Construct a frequency
distribution table of class interval 5 and find the median.

142, 145, 151, 157, 159, 160, 165, 162, 156, 158, 155, 141, 147, 149, 148, 159,
154, 155, 166, 168, 169, 172, 174, 173, 176, 161, 164, 163, 149, 150, 154, 153,
152, 164, 158, 159, 162, 157, 156, 155.

6. Work in group of 5. Collect the age of 50 students of grade play group to grade 12
randomly and construct a frequency distribution table of suitable interval.
Calculate the median of the age and present all process to the class.

17.3 Quartiles

Draw a line of length 12 cm or take a stick of length 12 inch. Mark a point on it such that

it is at equal distance from each end points. Again mark points on the parts so that they

are divided into two equal parts as shown in figure.

In this case, we can see that there are three A D C E B

points which divide the line/stick into four

equal parts. 0 3 6 9 12

We call them quartile. They are denoted by

Q1, Q2 and Q3 respectively. Also note that

Q2 is median, since it divides the distribution

into two equal parts.

Note: Q1 is called lower quartiles and Q3 is called upper quartile.
Calculation of quartiles

a) For individual series first arrange all data in ascending order and then use formula
( ) item for Q1 and ( ( )) item for Q3 to locate the value of quartiles.

b) For discrete series we have to use the following steps.

i. Construct less than cumulative frequency distribution table.

236 Mathematics, grade 10

ii. Use formula ( ) for Q1 and ( ) for Q3 to locate the quartiles and
find the value of quartiles in the column of X which is the corresponding
value of c.f. just greater than and ( ) respectively.

(c) To calculate quartiles Q1 and Q3 from continuous series, we have to follow the
following steps.
i. Construct cumulative frequency distribution table.

ii. Find the values of and for Q1 and Q3 respectively,
where N = ∑ = total frequency

iii. The corresponding class interval of value of or greater than in the

column of c.f for Q1 and or greater than in c.f for Q3 are called the
classes of Q1 and Q3 respectively.
iv. Use the formula

Q1 = L + xh

where L = Lower limit of class containing Q1,
c.f = Cumulative frequency of the class preceding the class containing Q1,
f = frequency of class containing Q1, and
h = length of class containing Q1.

Q3 = L+ xh

Where, L = Lower limit of class containing Q3
c.f = Cumulative frequency of the class preceding the class containing Q3
f = frequency of class containing Q3, and
h = length of class containing Q3
Example 1:
Calculate the values of Q1 and Q3 from the following data.

Age of workers 20 25 28 30 32 35 42 46

No of workers 2 8 12 10 14 7 5 1

Solution:
Construction of cumulative frequency data

Age (X) No of workers (f) c.f.

20 2 2

25 8 10

Mathematics, grade 10 237

28 12 22
30 10 32
32 14 46
35 7 53
42 5 58
46 1 59

f = N = 59

Here, the position of Q1 = ( ) item

= ( ) item

= 15 item
∴ The value of c.f equal to or just greater than 15 is 22.
∴ The corresponding value of c.f. 22 in X is Q1. i.e. Q1 = 28
Again, the position of Q3 = ( ( )) item

= ( ( )) item = 45 item

The value of c.f. equal to or just greater than 45 is 46.
∴ Q3 is the value of X corresponding to c.f. 46
i.e. Q3 = 32

Example 2:
Calculate the values of Q1 and Q3 from the following distribution table.

Mark obtained 10-20 20-30 30-40 40-50 50-60 60-70 70-80

No of students 2 8 15 14 10 8 3

Solution:
Calculation of quartiles,

Mark obtained frequency (f) Cumulative Frequency (c.f.)

10-20 2 2

20-30 8 10
30-40 15 25
40-50 14 39
50-60 10 49

238 Mathematics, grade 10

60-70 8 57

70-80 3 60

f = N = 60

Now, the position of Q1 = ( ) item

= ( ) item

=15 item
 Q1 lies in the interval 30-40 since 25 is just greater than 15 in column c.f
 L = 30, c.f = 10, f = 15 and h =10

Now, Q1 = L + xh

= 30 + x10

= 30 + 3.34 = 33.34

Again, the position of Q3 = item

= item

= 45 item
The value of c.f just greater than 45 is 49 in the column of c.f. So, Q3 lies in the class
50-60.
Where, L = 50, c.f= 39, f = 10 and h = 10

We have, Q3 = L+ xh

= 50+ x10

= 50 + 6 = 56

Exercise 17.4

1. Calculate the values of Q1 and Q3 from the following data.
(a) 10, 12, 14, 11, 22, 15, 27, 14, 16, 13, 25
(b) 250, 200, 150, 180, 190, 205, 208, 230, 155, 145, 149, 225, 202, 206, 257.
(c)

Mark 42 48 49 53 56 59 60 65 68 70

No of students 2 3 5 8 9 11 7 8 6 4

Mathematics, grade 10 239

(d)

Wages <200 210 215 220 225 230 >230

No. of workers 8 15 25 22 18 14 11

(e)

Marks <35 <40 <50 <55 <60 <65 <75 <85

No. of students 3 10 22 40 70 95 110 123

2. Calculate the values of Q1 and Q3 from the following data.
(a)

Age of students 2-4 4-6 6-8 8-10 10-12 12-14 14-16 16-18

No. of students 5 12 25 26 24 28 20 15

(b)

Mark obtained 10-20 20-30 30-40 40-50 50-60 60-70 70-80

No. of students 2 3 6 12 13 11 7

(c)

Height (cm) 100-110 110-120 120-130 130-140 140-150 150-160 160-170

No. of 3 4 9 15 20 14 7
students

(d)

Wages (Rs) 100-150 150-200 200-250 250-300 300-350 350-400

No. of 6 11 21 34 25 22
workers

(e)

Class 0-20 20-40 40-60 60-80 80-100 100-120 120-140

Frequency 8 12 15 14 12 9 10

3.(a) If Q1 = 8, find value of k in the following table.

Age (yr) 0-6 6-12 12-18 18-24 24-30 30-36

No. of persons 9 6 5 k 7 9

240 Mathematics, grade 10

(b) If Q1 = 31, find value of missing frequency in the following table.

Class 10-20 20-30 30-40 40-50 50-60 60-70
6
Frequency 4 5 ? 8 7

(c) If Q3 = 51.75, then find the value of k in the following table.

Weight (in kg) 40-44 44-48 48-52 52-56 56-60 60-64

Frequency 8 10 14 k 3 1

(d) What will be value of P if the upper quartile is Rs 460.

Income (Rs) 100-200 200-300 300-400 400-500 500-600

No. of person 15 18 P 20 17

4. Calculate the values of Q1 and Q3 from the following data
(a)

Height (cm) <125 <130 <135 <140 <145 <150 <155

No. of students 0 5 11 24 45 60 72

(b)

Weight 110- 120- 130- 140- 150- 160- 170- 180-
(kg) 119 129 139 149 159 169 179 189

No. of 5 7 12 20 16 10 7 3
students

5.(a) The marks obtained by 30 students are as follows.

42, 65, 78, 70, 62, 50, 72, 34, 30, 40, 58, 53, 30, 34, 51, 54, 42, 59, 20, 40, 42, 60,
25, 35, 35, 28, 46, 60

Construct a frequency distribution table of each class length 10 and find the value
of Q1 and Q3.

(b) Construct the class interval of length 20 and calculate lower and upper quartiles of
the following data.

32, 87, 17, 51, P9, 79, 64, 39, 25, 95, 53, 49, 78, 32, 42, 48, 59, 86, 69, 57, 15, 27,
44, 66, 77, 92.

6. Work in groups of 3 students. Collect the data of 100 students of your school
about the time required to reach the school from home. Present the data in the
frequency distribution table. Find the value that divides the whole data into four
equal classes and present your work to the class.

Mathematics, grade 10 241

17.4 Use of cumulative frequency curves (Ogives)

Cumulative frequency is useful if detailed information about the data distribution
is required. The curves of cumulative frequency are used to calculate the values of
quartiles and median. Mainly there are two types of cumulative frequency curves.
We call them as less than cumulative frequency curve and more than cumulative
frequency curve. (In other words, more than ogives and less then ogives.) The
point on X-axis corresponding to the point of intersection of more than
cumulative frequency curve and less than cumulative frequency curve is called the
median.

Example 1

Draw more than and less than cumulative curve of the following data

Height (cm) 90-100 100-110 110-120 120-130 130-140 140-150

frequency 5 22 30 31 18 6

Solution:
First construct the more than and less than frequency table as follow.

less than cumulative more than cumulative
frequency table frequency table

Height (cm) f Height (cm) Less than c.f. Height (cm) More than
c.f.

90-100 5 less than 100 5 more then 90 112

100-110 22 " " 110 27 " " 100 107

110-120 30 " " 120 57 " " 110 85

120-130 31 " " 130 88 " " 120 55

130-140 18 " " 140 106 " " 130 24

140-150 6 " " 150 112 " " 140 6

Now, for less than ogive, plot the points (100,5), (110,27), (120, 57) (130, 88), (140, 106)
and (150, 112) in graph and join the points without using scale.

Similarly, for more than ogive plot the points (90, 112), (100,107), (110, 85), (120, 55),
(130, 24), (140, 6) and join them without using scale. See the graph of these two ogives
in the following figures.

242 Mathematics, grade 10

Calculation of median, upper quartile and lower quartile by using cumulative
frequency curve:
The following steps should be completed to find partition value (Md, Q1, Q3) by
frequency distribution curves.
i. Find the position of Q1, Md and Q3 in Y-axis by using the formula , and

respectively.
ii. Draw horizontal line from a point obtained in Y-axis such that the line meets the

frequency curve.
iii. Draw vertical line from the point on the curve at which the horizontal line meet to

X - axis.

iv. The point at X- axis is our required value.

Example 2:

Compute Q1, Md, and Q3 from the given data by using graphic method.

Mark obtained 10-20 20-30 30-40 40-50 50-60 60-70 70-80

No. of students 2 8 15 14 10 8 3

Solution:
First construct less than frequency table.

marks less than cumulative frequency
less than 20 2
30 10
40 25
50 39
60 49
70 57

80 60

Mathematics, grade 10 243

Now, from less than ogive

i. Q1 lies on ( ) = ( ) = 15 item in Y-axis
So, the corresponding value of 15 in X-axis is 33.5. So first quartile is 33.5

ii. Median lies on ( ) =( ) = 30 item in Y-axis

So, the corresponding value of 30 in X-axis is 43.5 so median is
approximately 44.

iii. Q3 (upper quartile) lies on = 3x15 = 45th item in Y-axis Q3 lies in the

interval 50-60 and value in x axis corresponding to 45 is 56. Therefore,

Q3 = 56.

244 Mathematics, grade 10

Exercise 17.5
1. Draw less than ogive and find median class of the following data.

(a) Marks obtained 0-10 10-20 20-30 30-40 40-50 50- 60

No. of students 4 10 20 15 6 5

(b) Wages (Rs) 100- 150- 200- 250- 300- 350-
150 200 250 300 350 400

No. of workers 5 8 15 12 7 3

(c) Age of students 4-6 6-8 8-10 10-12 12-14 14-16 16-18
No. of students 7 12 21 15 14 11 10

(d) Expenses 5-10 10- 15- 20- 25- 30- 35- 40- 45-

(Rs.) 15 20 25 30 35 40 45 50

No. of 8 13 17 20 22 18 10 8 4
students

2. Construct less than ogive and more than ogive of the following data

(a) Marks obtained 20-30 30-40 40-50 50-60 60-70 70-80 80-90

No. of students 5 6 8 12 15 14 10

(b) Class interval 5-15 15-25 25-35 35-45 45-55 55-65
Frequency 5 12 30 10 8 5

(c) Wages (Rs) 20-40 40-60 60-80 80-100 100-120 120-140

No. of workers 4 5 10 8 7 6

(d) Weight of 40-44 40-48 40-52 40-56 40-60 40-64 40-68
students 10 34 49 60 67 10

No. of 9
students

Mathematics, grade 10 245

3. Calculate median class and value of median from the graph of question number 2.
4. Draw less than ogive of Q.N.1. and find the value of Q1 and Q3.
5. The daily expenses (in Rs) of 40 students of a class are given below.

6, 12, 35, 23, 65, 40, 37, 39, 28, 44, 32, 25, 18, 12, 9, 32, 55, 62, 49, 52, 26, 40, 32,
55, 14, 16, 20, 26, 54, 49, 50, 66, 68, 35, 42, 45, 39, 50, 24, 29.
Construct a frequency distribution of the class interval 10. After construct more
than and less than ogive.
Also find median, lower and upper quartiles by using graphical method.
6. Work in suitable group of students. Collect the data of 50 students about the
numbers of days of their parents' visit in the school per year. Construct frequency
distribution table with suitable length of interval. Construct less than and more
than ogive. Calculate the class and value of Q1, Q2, Q3 and than present to the
class.

246 Mathematics, grade 10